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The Feasibility of Testing LHVTs in High Energy Physics
李军利
中国科学院 研究生院
桂林 2006.10.27-11.01
In corporation with 乔从丰 教授
Phys.Rev. D74,076003, (2006)
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Content
EPR-B paradox. Bell inequality. Bell Inequality in Particle physics. The Feasibility of Testing LHVTs in Charm facto
ry.
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1.EPR-B paradox
In a complete theory there is an element corresponding to each element of reality.
Physical reality: possibility of predicting it with certainty, without disturbing the system.
Non-commuting operators are mutually incompatible.
I. The quantities correspond to non-commuting operators can not have simultaneously reality. or II. QM is incomplete.
Einstein, Podolsky, Rosen. 1935
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Two different measurements may performe upon the first particle. Due to angular momentum conservation and Einstein’s argument of reality and locality,the quantities of Non-commuting operators of the second particle can be simultaneously reality.
EPR: ( Bohm’s version)
So QM is incomplete !
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Bohr’s reply
Bohr contest not the EPR demonstration but the premises.
An element of reality is associated with a concretely performed act of measurement.
This makes the reality depend upon the process of measurement carried out on the first system.
That it is the theory which decides what is observable, not the other way around. ----Einstein
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2.Bell Inequality (BI)
EPR-(B) Hidden variable theory
Von Neuman (1932) : the hidden variable is unlikely to be true.Gleason(1957), Jauch(1963) , Kochen-Specher(1967) D>=3 paradox.
Bell D=2. Bell inequality(1964).D>=3. contextual dependent hidden variable theorem would survive.
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Hidden variable and Bell inequality
( , ) ( ) ( , ) ( , )E a b d A a B b
2 ( , ) - ( , ) ( , ) ( , ) 2S E a d E a b E c b E d c
ab
c
d
(a,b)= a bE a b QM:
Bell, physics I,195-200, 1964
CHSH, PRL23,880(1969)
LHVT:
1),(|),(),(|1 cbEbaEcaES
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Optical experiment and result
Aspect 1982 two channel polarizer.
PRL49,91(1982)2.697 0.015S
Experiment with pairs of photons produced with PDC
PRL81,3563(1998)
All these experiments conform the QM!
22)009.085.0( S W. Tittel, et al
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3.Bell Inequality in Particle physics Test BI with fermions or massive particles. Test BI with interactions other than electromagnet
ic interactions. Strong or Weak actions. Energy scale of photon case is eV range. Nonlocal
effects may well become apparent at length scale about cm.
1310 S.A. Abel et al. PLB 280,304 (1992)
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Bell Inequality in Particle physics
In spin system: the measurement of spin correlation in low-energy proton proton scattering. [M.Lamehi-Rachti,W.Mitting,PRD,14,2543,1976].
Spin singlet state particle decay to two spin one half particles. [N.A. Tornqvist. Found.Phys.,11,171,1981].
With meson system: Quasi spin system.
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}{1 00 KqKpN
K s
}{1 00 KqKpN
KL
}{2
1 001 KKK
}{2
1 002 KKK
Mass eigenstates CP eigenstates S eigenstates
00 KKS
00 KKS
Like the photon case they don’t commutate
}{2
11 0000, KKKK 0K 0K Are regard as the quasi-spin states.
)()cos(),( rl ttlr etmttE
2
iMH LSLSLS KKH SLSLLS
im
2
Berltamann, Quant-ph/0410028
Note that the H is not a observable [not hermitian]
Fix the quasi-spin and free in time.
1
2
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Experiment of 00BB system
A.Go. J.Mod.Opt.51,991.
lDBlDB *0*0 ,
They use 2S to test the BI:
as the flavor tag.
However, debates on whether it is a genuine test of LHVTs or not is still ongoing.
PLA332,355,(2004)
R.A. Bertlmann
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Other form of nonlocality
Nonlocality without using inequalities. GHZ states: three spin half particles.(1990)
Kochen-Specher: two spin one particles.(80)
L. Hardy: two spin one half particles.
Dimension-6
Hardy’s proof relies on a certain lack of symmetry of the entangled state.
PRL71, 1665 (1993)
.,
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GHZ states: three spin half particles.
)111000(2
1CBACBAABC
ABCABC
Cx
Bx
Ax
ABCABC
Cx
By
Ay
ABCABC
Cy
Bx
Ay
ABCABC
Cy
By
Ax
1
1
1
1
Cx
Bx
Ax
Cx
By
Ay
Cy
Bx
Ay
Cy
By
Ax
mmm
mmm
mmm
mmm
“No reasonable definition of reality could be expected to permit this.”
(1)
(2)
(3)
(4)
)3()2()1( 1)()()( 222 Cy
By
Ay
Cx
Bx
Ax mmmmmm
which contradict (4).1 Cx
Bx
Ax mmm
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Kochen-Specher: two spin one particles.
2222 zyx SSS(1). Any orthogonal frame (x, y, z), 0 happens exactly once.
(2). Any orthogonal pair (d, d’), 0 happens at most once.
)1,1,1()1,0,0(
)0,0,1()0,1,1()1,1,0(
)0,1,1()1,1,0()1,1,1(
76
543
210
aa
aaa
aaa
If h(a0)=h(a7)=0, then h(a1)=h(a2)=h(a3)=h(a4)=1.
So that h(a5)=h(a6)=0, by (1), which contradict (2).
A set of eight directions represented in following graph:
Consider a pair of spin 1 particle in singlet state. So can determine the value for Si without disturbing that system, leading the non-contextuality.
R.A. Bertlmann & A. Zeilinger quantum [un]speakables from Bell to quantum information
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L. Hardy: two spin one half particles.
0.1 FG 0)1(.2 GD
0)1(.3 EF 0.4 DE
But because of 2 & 3. If D=1 then G=1. If E=1 then F=1.
So if the probability that D=E=1 is not 0,
then the probability for F=G=1 won’t be 0.
Jordan proved that for the state like:
PRA50, 62 (1994) Jordan
mep i sincos
1tan If there exist four projection operators satisfy:
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)]'()['()'()()()]([)()'( bBIaAbBaAbBaAIbBaA
1 234
0.1 FG 0)1(.2 GD
0)1(.3 EF 0.4 DE
Compare to previous page:
0)(*,,*)'()','(),'()',(),(1 bPaPbaPbaPbaPbaP
0)()'()'()'(
)()'()'()()()(1
bIBIaAbBaA
bBaAbBaAbBaA
Consider the CH inequality: PRD10, 526 (1974)
PRA52, 2535 (1995) Garuccio
Eberhard Inequality
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Hardy type experiment with entangled Kaon pairs Generate a asymmetric state. Eberhard’s inequality (EI).
][2
1)(
2 LLSLLS KRKKKKKR
T
)()()()( 0000 KKPKKPKKPKKP LLRSSLRLLRLR
PRL88, 040403 (2002),
PRL89, 160401 (2002).
Quant-ph/05011069.
A. Bramon, and G. Garbarino:
A. Bramon, R. Escribano and G. Garbarino:
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][2
1)( LLSSSLLS KrKKrKKKKKt
To generate the asymmetric state, fix a thin regenerator on the right beam close to decay points. Then the initial state:
][2
1SLLS KKKK
Becomes:
Let this state propagate to a proper time T:
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Normalize it to the surviving pairs leads to:
][2
12 LLSLLS KRKKKKK
R
.Re 2)( TTmmii
LSSL
reR
where
]
[2
)()(
)(2
1)(
2
1
SS
TmTiLL
TmTi
SLLS
KKereKKere
KKKKTN
T
SLLS
component has been enhanced.
has been further suppressed.
LLKK
SS KK
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4.The Feasibility of Testing LHVTs in Charm factory Easy to get space-like separation. Can test the phenomena: less entangled state
leads to larger violation of inequality.
][2
1)(/ 2
)(
2 LL
TTmmi
SLLS KKreKKKKR
TJLS
SL
][2
1)0(/ 0000 KKKKJ
In the charm factory the entanglement state formed as:
.Re 2)( TTmmii
LSSL
re
where
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)2(4
Re2),( 2
2
00
RKKP
i
QM
)2(2
Re1),( 2
2
0
RKKP
i
LQM
)2(2
Re1),( 2
2
0
RKKP
i
LQM
0),( SSQM KKP
The four joint measurement of the transition probability needed in the EI predicted by QM take the following form:
20000 )(/),( TJKKKKPQM
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.)2(2
)1(0
)2(2
)1(
)2(4
)2(2
2
2
2
2
2
R
R
R
R
R
R
Take into EI:
)]()()([)( 0000 KKPKKPKKPKKPVD LLRSSLRLLRLREI
See Figure 1
0 0(for See figure 3)
where VDis the violation degree of the inequality.
From QM we have: First assume
.)2(4
432
2
R
RRVDEI
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Actually has non-zero magnitude :
The shaded region is the requirement of the real and imagine part of R when violation between QM and LHVTs can be seen from inequality.
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The advantage of over
22.0K
94.0K
factory
/J Charm factory
Space-like separation required:
/J
4.010 632
eerRTS
31.032 1010
eerRTS
11 ST In
To make sure the misidentification of is of order per thousandS LK with K
ST 5Properly choose PRL88, 040403 (2002)
0.15 ST
4.0R
310R
so
so
has a wider region of R in discriminating QM from LHVT./J
There is phenomena can be test due to this advantage.
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Quantify the entanglement
TLSerRJJJC
)(222
2
2
2~/
~/)/(
PRL80, 2245 (1998)W. Wootters*21 )/(
~/
~ JJ yyWhere:
This mean the state become less entangled during time evolution!
Historically the amount of the violation was seen as extent of entanglement. This may not be the case in EI. As indicated in Figure 1 & 2.
To see this we must quantify the degree of entanglement Take concurrence as a measure of this quantity.
C changes between 0 to 1 for no entanglement and full entanglement.
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2 ( , ) - ( , ) ( , ) ( , ) 2S E a d E a b E c b E d c
)1(22 CS
Express the violation in degree of entanglement
PLA154,201(1991)
Abouraddy et al.
N.Gisin
PRA64,050101,(2001)
)2(4
432
2
R
RRVDEI
2
2
2
RC
4
22)1(3 2CCCVDEI
1.The usual CHSH inequality:
2. The Hardy state using Eberhard’s inequality :See the figure next page
Note we make a trick in the figure that substitute C with .1 2x
2)1(2 CVDCHSH
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The Entanglement and Bell inequality violation
Magnitudes below zero of VD is the range of violation
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Thank you for your patience.